The angle of elevation of the top of a pole from a point A on the ground is $$30^\circ$$. The angle of elevation changes to $$45^\circ$$, after moving 20 meters towards the base of the pole. Then the height of the pole, in meters, is

Since $$\triangle ACD$$ has $$\angle ADC =\angle CAD =45^\circ$$, it is an isosceles right-angled triangle, and the lengths $$DC$$ and $$AD$$ will both be the same. $$DC$$ being the length of the pole, we find $$DC=AD= p$$ meters.

Since $$AB$$ is the length walked, it is $$20$$ meters, and the total length $$BC$$ is $$p+20$$ meters.

In $$\triangle BCD$$, since the pole is upright, $$\angle DCB$$ is $$90^\circ$$, and the other two angles are $$30^\circ$$ and $$60^\circ$$. We therefore get it as a $$30-60-90$$ triangle and the ratio of sides $$1:\sqrt{3}:2$$ will give $$DC:BC = 1:\sqrt{3}$$.

We finally have $$\dfrac{p}{p+20}=\dfrac{1}{\sqrt{3}}$$ which gives $$p=\dfrac{20}{\sqrt{3}-1} = 10(\sqrt{3}+1)$$